Finding x-intercepts and vertexes of quadratics.

ok, so i'm having a rough time trying to get an answer, mostly because i don't have confidence on my answers. (i already answered question #1)

2) m^2 = 9m-1

for this, i rearranged it to become m^2-9m+1=0 (in accordance w/the standard form i think). so my vertex is (4.5, 61.75) and the x-intercepts i got were {0.1125, 8.8875}. my graph is a parabola pointing downward...

3) -25=x^2+10x

for this, i rearranged to get y=x^2+10x+25. i got a vertex of (-5,0). i'm stuck here because i'm not even sure what i did in #2 was correct.
* edit * i tried to get x-intercepts and i got two -5s. i really don't think my x-intercepts are right.

ok, so i'm having a rough time trying to get an answer, mostly because i don't have confidence on my answers. (i already answered question #1)

2) m^2 = 9m-1

for this, i rearranged it to become m^2-9m+1=0 (in accordance w/the standard form i think). so my vertex is (4.5, 61.75) and the x-intercepts i got were {0.1125, 8.8875}. my graph is a parabola pointing downward...

intercepts are right, vertex is wrong. i think you miscalculated when you pluged in the 4.5

3) -25=x^2+10x

for this, i rearranged to get y=x^2+10x+25. i got a vertex of (-5,0). i'm stuck here because i'm not even sure what i did in #2 was correct.
* edit * i tried to get x-intercepts and i got two -5s. i really don't think my x-intercepts are right.

thanks in advance.

you are correct. the vertex happens to be the only x-intercept. the vertex touches the x-axis, so the x-intercept is one point, namely the vertex. trust yourself more

i thought that because i thought everything had to have coordinates, lol.

umm, every point on the graph does have a coordinate.

don't let the fact that the parabola has one x-intercept mess you up, there are some with no x-intercept, it just means the parabola always returns values that are positive, but it always returns values, since parabolas are defined for all x.